Why is it impossible to have a 3 by 3 magic square with the numbers 1, 3, 4, 5, 6, 7, 8, 9, and 10?

Question

Why is it impossible to have a 3 by 3 magic square with the numbers 1, 3, 4, 5, 6, 7, 8, 9, and 10?

Answer 1

Those numbers add up to 53 which is a prime number

and therefore has no factors, other than 1 and 53.

The sum of the numbers needs to be divisible by 3, as

they have to be distributed into three rows (or columns)

with each row (or column) having the same sum.

Answer 2

Magic square is a square that the sum of each column and each row are equal. (or maybe the diagonal too, but we'll see that each row and each column have the same sum is imposibble). Let's say its sum as S.

Then row1 + row2 + row3 + column1 + column2 + column3 = 6S

Let's say that we've put the numbers like this

abc

def

ghi

note that (a+b+c+d+e+f+g+h+i) = (1+3+4+5+6+7+8+9+10) = 53

then, row1 + row2 + row3 + column1 + column2 + column3 = a+b+c + d+e+f + g+h+i + a+d+g + b+e+h + c+f+i = 2a + 2b + 2c + 2d + 2e + 2f + 2g + 2h + 2i = 2(a+b+c+d+e+f+g+h+i) = 2 . 53 = 106

Then 6S = 106 => S = 17.5

Because S should be an integer, this is imposibble.